find the fundamental frequency and the frequency of the first two overtones of a pipe `45.0 cm` long. (a) If the pipe is open at both ends. (b) If the pipe is closed at one end. Use `upsilon = 344 m//s` .

Explanation

To solve the problem of finding the fundamental frequency and the frequencies of the first two overtones of a pipe that is 45.0 cm long, we will consider two cases: (a) when the pipe is open at both ends and (b) when the pipe is closed at one end. The speed of sound (ν) is given as 344 m/s.

### Part (a): Pipe Open at Both Ends

1. **Convert Length to Meters**:
- Given length \( L = 45.0 \, \text{cm} \).
- Convert to meters:
\[ L = \frac{45.0}{100} = 0.45 \, \text{m} \]

2. **Calculate the Fundamental Frequency**:
- The formula for the fundamental frequency \( f_1 \) of an open pipe is:
\[ f_1 = \frac{V}{2L} \]
- Substitute the values:
\[ f_1 = \frac{344}{2 \times 0.45} = \frac{344}{0.9} \approx 382.2 \, \text{Hz} \]

3. **Calculate the First Overtone**:
- The first overtone \( f_2 \) for an open pipe is:
\[ f_2 = \frac{2V}{2L} = \frac{V}{L} \]
- Substitute the values:
\[ f_2 = \frac{344}{0.45} \approx 764.4 \, \text{Hz} \]

4. **Calculate the Second Overtone**:
- The second overtone \( f_3 \) for an open pipe is:
\[ f_3 = \frac{3V}{2L} \]
- Substitute the values:
\[ f_3 = \frac{3 \times 344}{2 \times 0.45} = \frac{1032}{0.9} \approx 1146.6 \, \text{Hz} \]

### Part (b): Pipe Closed at One End

1. **Calculate the Fundamental Frequency**:
- The formula for the fundamental frequency \( f_1 \) of a closed pipe is:
\[ f_1 = \frac{V}{4L} \]
- Substitute the values:
\[ f_1 = \frac{344}{4 \times 0.45} = \frac{344}{1.8} \approx 191.1 \, \text{Hz} \]

2. **Calculate the First Overtone**:
- The first overtone \( f_2 \) for a closed pipe is:
\[ f_2 = \frac{3V}{4L} \]
- Substitute the values:
\[ f_2 = \frac{3 \times 344}{4 \times 0.45} = \frac{1032}{1.8} \approx 573.0 \, \text{Hz} \]

3. **Calculate the Second Overtone**:
- The second overtone \( f_3 \) for a closed pipe is:
\[ f_3 = \frac{5V}{4L} \]
- Substitute the values:
\[ f_3 = \frac{5 \times 344}{4 \times 0.45} = \frac{1720}{1.8} \approx 955.5 \, \text{Hz} \]

### Summary of Results:
- For the pipe open at both ends:
- Fundamental frequency: \( 382.2 \, \text{Hz} \)
- First overtone: \( 764.4 \, \text{Hz} \)
- Second overtone: \( 1146.6 \, \text{Hz} \)

- For the pipe closed at one end:
- Fundamental frequency: \( 191.1 \, \text{Hz} \)
- First overtone: \( 573.0 \, \text{Hz} \)
- Second overtone: \( 955.5 \, \text{Hz} \)